Is there a way to find the radical ideal of $I_i=(a^n-u^{n-i+1}v^{n-i}, b^n-u^{i-1}v^i, uv-ab)$ for $1\leq i \leq n$ in $\mathbb{C}[u,v,a,b]?$

This is the generalization of my question here where I wanted to use Macaulay2 software to compute the radical ideal for $n=3$ and $i=2$ of the above ideal. Unfortunately, I don't know how to use the software in the general case, and I don't know if it works or not. At least, using Macaulay2 for some special cases I can guess that $\sqrt{I_i}=(a^i-u^{n-i}b^{i-1}, b^{n-i+1}-v^{i-1}a^{n-i}, uv-ab)$ but there is problem in my further computation, so I thought maybe what I guessed is wrong!

I would appreciate any help on that.

Motivation:

This is indeed related to the Derived McKay correspondence where I'm studying the image of torus-invariant, zero-dimensional sheaves of the minimal resolution $Y$ of $\mathbb{C}^2/\mathbb{Z}/n$ under the Fourier-Mukai transform from the (bounded) derived category of coherent sheaves on $Y$ to the (bounded) derived category of coherent sheaves on $\mathfrak{X}=[\mathbb{C}^2/\mathbb{Z}/n],$ the stacky resolution of $\mathbb{C}^2/\mathbb{Z}/n.$

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Could you elaborate concerning the problem in your further computation? – JSchlather Jan 22 '13 at 1:00
It would be helpful if you could provide the code you are using to compute the radical of the ideal, it is likely that you are not getting the answer you expect due to a bug in your algorithm. – Charles Boyd Jan 22 '13 at 1:09
Dear @Jacob Schlather, I added a little bit motivation! – Ehsan M. Kermani Jan 22 '13 at 1:38
Dear @Charles Boyd, I used the same code as Fredrik Meyer provided in the given link above, for the general case, but I don't know, it simply doesn't work. – Ehsan M. Kermani Jan 22 '13 at 1:42

Is this roughly what you want to do?

Script: ~/tmp/ideals.m2

R = QQ[u,v,a,b];

n = 10;

myIdeal = (i,n) -> (
p1 = a^n - (u^(n-i+1))*(v^(n-i));
p2 = b^n - (u^(i-1))*(v^i);
p3 = u*v - a*b;
ideal(p1,p2,p3)
)

)

for j from 1 to n do (
J = myIdeal(j,n);
print rJ
)


Output: (for $1 \leq i \leq 10$)

ii60 : load "~/tmp/ideals.m2"
10           9
ideal (- u*v + a*b, b   - v, - u*b  + a)
9             8    2
ideal (- u*v + a*b, b  - v*a, - u*b  + a )
8      2       7    3
ideal (- u*v + a*b, b  - v*a , - u*b  + a )
7      3       6    4
ideal (- u*v + a*b, b  - v*a , - u*b  + a )
6      4       5    5
ideal (- u*v + a*b, b  - v*a , - u*b  + a )
6      4       5    5
ideal (- u*v + a*b, a  - u*b , - v*a  + b )
7      3       6    4
ideal (- u*v + a*b, a  - u*b , - v*a  + b )
8      2       7    3
ideal (- u*v + a*b, a  - u*b , - v*a  + b )
9             8    2
ideal (- u*v + a*b, a  - u*b, - v*a  + b )
10           9
ideal (- u*v + a*b, a   - u, - v*a  + b)


Example: ($i=2$, $n=3$)

ii61 : J2 = myIdeal(2,3)

2     3       2    3
oo61 = ideal (- u v + a , - u*v  + b , u*v - a*b)

oo61 : Ideal of R

2               2
oo62 = ideal (a  - u*b, v*a - b , u*v - a*b)

oo62 : Ideal of R

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Yes indeed, dear @Charles. Thank you very much for your time. That helps a lot. – Ehsan M. Kermani Jan 22 '13 at 5:10
The function myRad is not necessary, it is just equal to the function radical. – Thomas Jun 27 '15 at 8:47

To compute the radical of a binomial ideal, in some cases it can be advantageous (that is, faster) to use the Binomials package in Macaulay2. Here is the documentation. In Charles Boyd's answer you can just replace the call to radical by binomialRadical.

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